Capacitance and Dielectrics - mytextbookswheniwantem · PDF fileCapacitance and Dielectrics...

740

26.1 Definition of Capacitance

26.2 Calculating Capacitance

26.3 Combinations of Capacitors

26.4 Energy Stored in a Charged Capacitor

26.5 Capacitors with Dielectrics

26.6 Electric Dipole in an Electric Field

26.7 An Atomic Description of Dielectrics

In this chapter, we introduce the first of three simple

circuit elements that can be connected with wires to form

an electric circuit. Electric circuits are the basis for the vast

majority of the devices used in our society. Here we shall

discuss capacitors, devices that store electric charge. This dis-

cussion is followed by the study of resistors in Chapter 27 and

inductors in Chapter 32. In later chapters, we will study more

sophisticated circuit elements such as diodes and transistors.

Capacitors are commonly used in a variety of electric cir-

cuits. For instance, they are used to tune the frequency of

radio receivers, as filters in power supplies, to eliminate spark-

ing in automobile ignition systems, and as energy-storing

devices in electronic flash units.

chapter 26Capacitance and Dielectrics

26.1 Definition of CapacitanceConsider two conductors as shown in Figure 26.1. Such a combination of two con-ductors is called a capacitor. The conductors are called plates. If the conductors carry charges of equal magnitude and opposite sign, a potential difference DV exists between them. What determines how much charge is on the plates of a capacitor for a given volt-age? Experiments show that the quantity of charge Q on a capacitor1 is linearly pro-

1Although the total charge on the capacitor is zero (because there is as much excess positive charge on one conduc-tor as there is excess negative charge on the other), it is common practice to refer to the magnitude of the charge on either conductor as “the charge on the capacitor.”

When a patient receives a shock from a defibrillator, the energy delivered to the patient is initially stored in a capacitor. We will study capacitors and capacitance in this chapter. (Andrew Olney/Getty Images)

26.1 | Definition of Capacitance 741

portional to the potential difference between the conductors; that is, Q ~ DV. The proportionality constant depends on the shape and separation of the conductors.2

This relationship can be written as Q 5 C DV if we define capacitance as follows:

The capacitance C of a capacitor is defined as the ratio of the magnitude of the charge on either conductor to the magnitude of the potential difference between the conductors:

C ;Q

DV (26.1)

By definition capacitance is always a positive quantity. Furthermore, the charge Q and the potential difference DV are always expressed in Equation 26.1 as positive quantities. From Equation 26.1, we see that capacitance has SI units of coulombs per volt. Named in honor of Michael Faraday, the SI unit of capacitance is the farad (F):

1 F 5 1 C/V

The farad is a very large unit of capacitance. In practice, typical devices have capac-itances ranging from microfarads (1026 F) to picofarads (10212 F). We shall use the symbol mF to represent microfarads. In practice, to avoid the use of Greek letters, physical capacitors are often labeled “mF” for microfarads and “mmF” for micromi-crofarads or, equivalently, “pF” for picofarads. Let’s consider a capacitor formed from a pair of parallel plates as shown in Figure 26.2. Each plate is connected to one terminal of a battery, which acts as a source of potential difference. If the capacitor is initially uncharged, the battery establishes an electric field in the connecting wires when the connections are made. Let’s focus on the plate connected to the negative terminal of the battery. The electric field in the wire applies a force on electrons in the wire immediately outside this plate; this force causes the electrons to move onto the plate. The movement continues until the plate, the wire, and the terminal are all at the same electric potential. Once this equilibrium situation is attained, a potential difference no longer exists between the terminal and the plate; as a result, no electric field is present in the wire and the electrons stop moving. The plate now carries a negative charge. A similar pro-cess occurs at the other capacitor plate, where electrons move from the plate to the wire, leaving the plate positively charged. In this final configuration, the potential difference across the capacitor plates is the same as that between the terminals of the battery.

Definition of capacitance

Pitfall Prevention 26.1Capacitance Is a CapacityTo understand capacitance, think of similar notions that use a similar word. The capacity of a milk carton is the volume of milk it can store. The heat capacity of an object is the amount of energy an object can store per unit of temperature difference. The capacitance of a capacitor is the amount of charge the capacitor can store per unit of potential difference.

2The proportionality between Q and DV can be proven from Coulomb’s law or by experiment.

Pitfall Prevention 26.2Potential Difference Is DV, Not VWe use the symbol DV for the poten-tial difference across a circuit ele-ment or a device because this nota-tion is consistent with our definition of potential difference and with the meaning of the delta sign. It is a common but confusing practice to use the symbol V without the delta sign for both a potential and a poten-tial difference! Keep that in mind if you consult other texts.

�Q

�Q

When the capacitor is charged, the conductors carry charges of equal magnitude and opposite sign.

Figure 26.1 A capacitor consists of two conductors.

d

�Q�Q

Area � A

� �

When the capacitor is connected to the terminals of a battery, electrons transfer between the plates and the wires so that the plates become charged.

Figure 26.2 A parallel-plate capac-itor consists of two parallel conduct-ing plates, each of area A, separated by a distance d.

742 CHAPTER 26 | Capacitance and Dielectrics

Quick Quiz 26.1 A capacitor stores charge Q at a potential difference DV. What happens if the voltage applied to the capacitor by a battery is doubled to 2 DV ? (a) The capacitance falls to half its initial value, and the charge remains the same. (b) The capacitance and the charge both fall to half their initial values. (c) The capacitance and the charge both double. (d) The capacitance remains the same, and the charge doubles.

26.2 Calculating CapacitanceWe can derive an expression for the capacitance of a pair of oppositely charged conductors having a charge of magnitude Q in the following manner. First we cal-culate the potential difference using the techniques described in Chapter 25. We then use the expression C 5 Q/DV to evaluate the capacitance. The calculation is relatively easy if the geometry of the capacitor is simple. Although the most common situation is that of two conductors, a single conduc-tor also has a capacitance. For example, imagine a spherical, charged conductor. The electric field lines around this conductor are exactly the same as if there were a conducting, spherical shell of infinite radius, concentric with the sphere and car-rying a charge of the same magnitude but opposite sign. Therefore, we can identify the imaginary shell as the second conductor of a two-conductor capacitor. The elec-tric potential of the sphere of radius a is simply keQ /a (see Section 25.6), and setting V 5 0 for the infinitely large shell gives

C 5Q

DV5

Q

k eQ /a5

ak e

5 4pP0a (26.2)

This expression shows that the capacitance of an isolated, charged sphere is pro-portional to its radius and is independent of both the charge on the sphere and its potential. The capacitance of a pair of conductors is illustrated below with three familiar geometries, namely, parallel plates, concentric cylinders, and concentric spheres. In these calculations, we assume the charged conductors are separated by a vacuum.

Parallel-Plate Capacitors

Two parallel, metallic plates of equal area A are separated by a distance d as shown in Figure 26.2. One plate carries a charge 1Q, and the other carries a charge 2Q. The surface charge density on each plate is s 5 Q /A. If the plates are very close together (in comparison with their length and width), we can assume the electric field is uniform between the plates and zero elsewhere. According to the What If? feature of Example 24.5, the value of the electric field between the plates is

E 5s

P05

Q

P0A

Because the field between the plates is uniform, the magnitude of the potential dif-ference between the plates equals Ed (see Eq. 25.6); therefore,

DV 5 Ed 5Qd

P0A

Substituting this result into Equation 26.1, we find that the capacitance is

C 5Q

DV5

Q

Qd/P0A

C 5P0Ad

(26.3)

That is, the capacitance of a parallel-plate capacitor is proportional to the area of its plates and inversely proportional to the plate separation.

Capacitance of an isolated charged sphere

Capacitance of parallel plates

Pitfall Prevention 26.3Too Many CsDo not confuse an italic C for capaci-tance with a nonitalic C for the unit coulomb.

Example 26.1 The Cylindrical Capacitor

A solid cylindrical conductor of radius a and charge Q is coaxial with a cylindrical shell of negligible thick-ness, radius b . a, and charge 2Q (Fig. 26.4a). Find the capacitance of this cylindrical capacitor if its length is ,.

SOLUTION

Conceptualize Recall that any pair of conductors quali-fies as a capacitor, so the system described in this exam-ple therefore qualifies. Figure 26.4b helps visualize the electric field between the conductors.

Categorize Because of the cylindrical symmetry of the system, we can use results from previous studies of cylin-drical systems to find the capacitance.

Analyze Assuming , is much greater than a and b, we can neglect end effects. In this case, the electric field is perpendicular to the long axis of the cylinders and is confined to the region between them (Fig. 26.4b).

26.2 | Calculating Capacitance 743

Let’s consider how the geometry of these conductors influences the capacity of the pair of plates to store charge. As a capacitor is being charged by a battery, elec-trons flow into the negative plate and out of the positive plate. If the capacitor plates are large, the accumulated charges are able to distribute themselves over a substantial area and the amount of charge that can be stored on a plate for a given potential difference increases as the plate area is increased. Therefore, it is reason-able that the capacitance is proportional to the plate area A as in Equation 26.3. Now consider the region that separates the plates. Imagine moving the plates closer together. Consider the situation before any charges have had a chance to move in response to this change. Because no charges have moved, the electric field between the plates has the same value but extends over a shorter distance. There-fore, the magnitude of the potential difference between the plates DV 5 Ed (Eq. 25.6) is smaller. The difference between this new capacitor voltage and the termi-nal voltage of the battery appears as a potential difference across the wires connect-ing the battery to the capacitor, resulting in an electric field in the wires that drives more charge onto the plates and increases the potential difference between the plates. When the potential difference between the plates again matches that of the battery, the flow of charge stops. Therefore, moving the plates closer together causes the charge on the capacitor to increase. If d is increased, the charge decreases. As a result, the inverse relationship between C and d in Equation 26.3 is reasonable.

Quick Quiz 26.2 Many computer keyboard buttons are constructed of capaci-tors as shown in Figure 26.3. When a key is pushed down, the soft insulator between the movable plate and the fixed plate is compressed. When the key is pressed, what happens to the capacitance? (a) It increases. (b) It decreases. (c) It changes in a way you cannot determine because the electric circuit con-nected to the keyboard button may cause a change in DV.

KeyB

Movable plate

InsulatorFixed plate

Figure 26.3 (Quick Quiz 26.2) One type of computer keyboard button.

ba

�

Gaussiansurface

�Q

�Q

aQ

Q

b

r

a b

Figure 26.4 (Example 26.1) (a) A cylindrical capacitor consists of a solid cylindrical conductor of radius a and length , sur-rounded by a coaxial cylindrical shell of radius b. (b) End view. The electric field lines are radial. The dashed line represents the end of a cylindrical gaussian surface of radius r and length ,.

continuedWrite an expression for the potential difference between the two cylinders from Equation 25.3:

Vb 2 Va 5 2 3b

a ES? d sS


26.1 cont.

Example 26.2 The Spherical Capacitor

A spherical capacitor consists of a spherical conducting shell of radius b and charge 2Q concentric with a smaller conducting sphere of radius a and charge Q (Fig. 26.5). Find the capacitance of this device.

SOLUTION

Conceptualize As with Example 26.1, this system involves a pair of conductors and qualifies as a capacitor.

Categorize Because of the spherical symmetry of the sys-tem, we can use results from previous studies of spheri-cal systems to find the capacitance.

Apply Equation 24.7 for the electric field outside a cylin-drically symmetric charge distribution and notice from Figure 26.4b that E

S is parallel to d sS along a radial line:

Vb 2 Va 5 2 3b

a Erdr 5 22ke l 3

b

a

drr

5 22ke l ln a bab

Substitute the absolute value of DV into Equation 26.1 and use l 5 Q /,:

C 5Q

DV5

Q12k eQ /, 2 ln 1b/a 2 5

,

2k e ln 1b/a 2 (26.4)

Finalize The capacitance is proportional to the length of the cylinders. As you might expect, the capacitance also depends on the radii of the two cylindrical conductors. Equation 26.4 shows that the capacitance per unit length of a combination of concentric cylindrical conductors is

C,

51

2k e ln 1b/a 2 (26.5)

An example of this type of geometric arrangement is a coaxial cable, which consists of two concentric cylindrical conduc-tors separated by an insulator. You probably have a coaxial cable attached to your television set or video recorder if you are a subscriber to cable television. The coaxial cable is especially useful for shielding electrical signals from any possible external influences.

WHAT IF? Suppose b 5 2.00a for the cylindrical capacitor. You would like to increase the capacitance, and you can do so by choosing to increase either , by 10% or a by 10%. Which choice is more effective at increasing the capacitance?

Answer According to Equation 26.4, C is proportional to ,, so increasing , by 10% results in a 10% increase in C. For the result of the change in a, let’s use Equation 26.4 to set up a ratio of the capacitance C9 for the enlarged cylinder radius a9 to the original capacitance:

C rC

5,/2k e ln 1b/a r 2,/2k e ln 1b/a 2 5

ln 1b/a 2ln 1b/a r 2

We now substitute b 5 2.00a and a9 5 1.10a, representing a 10% increase in a:

C rC

5ln 12.00a/a 2

ln 12.00a/1.10a 2 5ln 2.00ln 1.82

5 1.16

which corresponds to a 16% increase in capacitance. Therefore, it is more effective to increase a than to increase ,. Note two more extensions of this problem. First, it is advantageous to increase a only for a range of relationships between a and b. If b . 2.85a, increasing , by 10% is more effective than increasing a (see Problem 66). Second, if b decreases, the capacitance increases. Increasing a or decreasing b has the effect of bringing the plates closer together, which increases the capacitance.

a

b

�Q

�Q

Figure 26.5 (Example 26.2) A spherical capacitor consists of an inner sphere of radius a sur-rounded by a concentric spherical shell of radius b. The electric field between the spheres is directed radially outward when the inner sphere is positively charged.

26.3 | Combinations of Capacitors 745

26.3 Combinations of CapacitorsTwo or more capacitors often are combined in electric circuits. We can calculate the equivalent capacitance of certain combinations using methods described in this section. Throughout this section, we assume the capacitors to be combined are initially uncharged. In studying electric circuits, we use a simplified pictorial representation called a circuit diagram. Such a diagram uses circuit symbols to represent various cir-cuit elements. The circuit symbols are connected by straight lines that represent the wires between the circuit elements. The circuit symbols for capacitors, batter-ies, and switches as well as the color codes used for them in this text are given in Figure 26.6. The symbol for the capacitor reflects the geometry of the most com-mon model for a capacitor, a pair of parallel plates. The positive terminal of the battery is at the higher potential and is represented in the circuit symbol by the longer line.

Parallel Combination

Two capacitors connected as shown in Active Figure 26.7a (page 746) are known as a parallel combination of capacitors. Active Figure 26.7b shows a circuit diagram for this combination of capacitors. The left plates of the capacitors are connected to the positive terminal of the battery by a conducting wire and are therefore both at the same electric potential as the positive terminal. Likewise, the right plates are connected to the negative terminal and so are both at the same potential as the negative terminal. Therefore, the individual potential differences across capacitors

26.2 cont.

Analyze As shown in Chapter 24, the direction of the electric field outside a spherically symmetric charge distribution is radial and its magnitude is given by the expression E 5 keQ /r 2. In this case, this result applies to the field between the spheres (a , r , b).

Substitute the absolute value of DV into Equation 26.1: C 5Q

DV5

Q0 Vb 2 Va 0

5ab

k e 1b 2 a 2 (26.6)

Apply the result of Example 24.3 for the electric field outside a spherically symmetric charge distribution and note that E

S is parallel to d sS along a radial line:

Vb 2 Va 5 2 3b

a Er dr 5 2k eQ 3

b

a

drr 2 5 k eQ c1r d

b

a

(1) Vb 2 Va 5 k eQ a1b

21ab 5 k eQ

a 2 bab

Write an expression for the potential difference between the two conductors from Equation 25.3:

Vb 2 Va 5 2 3b

a ES? d sS

Finalize The potential difference between the spheres in Equation (1) is negative because Q is positive and b . a. There-fore, in Equation 26.6, when we take the absolute value, we change a 2 b to b 2 a. The result is a positive number.

WHAT IF? If the radius b of the outer sphere approaches infinity, what does the capacitance become?

Answer In Equation 26.6, we let b S `:

C 5 limbS`

ab

k e 1b 2 a 2 5ab

k e 1b 25

ak e

5 4pP0a

Notice that this expression is the same as Equation 26.2, the capacitance of an isolated spherical conductor.

Capacitorsymbol

Batterysymbol

symbolSwitch Open

Closed

�

�

Figure 26.6 Circuit symbols for capacitors, batteries, and switches. Notice that capacitors are in blue, batteries are in green, and switches are in red. The closed switch can carry current, whereas the open one cannot.


connected in parallel are the same and are equal to the potential difference applied across the combination. That is,

DV1 5 DV2 5 DV

where DV is the battery terminal voltage. After the battery is attached to the circuit, the capacitors quickly reach their maximum charge. Let’s call the maximum charges on the two capacitors Q 1 and Q 2. The total charge Q tot stored by the two capacitors is the sum of the charges on the individual capacitors:

Q tot 5 Q 1 1 Q 2 (26.7)

Suppose you wish to replace these two capacitors by one equivalent capacitor hav-ing a capacitance Ceq as in Active Figure 26.7c. The effect this equivalent capacitor has on the circuit must be exactly the same as the effect of the combination of the two individual capacitors. That is, the equivalent capacitor must store charge Q tot when connected to the battery. Active Figure 26.7c shows that the voltage across the equivalent capacitor is DV because the equivalent capacitor is connected directly across the battery terminals. Therefore, for the equivalent capacitor,

Q tot 5 C eq DV

Substituting for the charges in Equation 26.7 gives

C eq DV 5 Q 1 1 Q 2 5 C 1 DV1 1 C 2 DV2

C eq 5 C 1 1 C 2 1parallel combination 2 where we have canceled the voltages because they are all the same. If this treat-ment is extended to three or more capacitors connected in parallel, the equivalent capacitance is found to be

C eq 5 C 1 1 C 2 1 C 3 1 c 1parallel combination 2 (26.8)

Therefore, the equivalent capacitance of a parallel combination of capacitors is (1) the algebraic sum of the individual capacitances and (2) greater than any of the individual capacitances. Statement (2) makes sense because we are essentially com-bining the areas of all the capacitor plates when they are connected with conduct-ing wire, and capacitance of parallel plates is proportional to area (Eq. 26.3).

Equivalent capacitance for capacitors in parallel

C2

C1

V

Q2

C2

Q1

C1

VV� �

V1�

�

� ��

�Q 1 �Q 1

V2�

�Q 2 �Q 2

� �

� �

� �

Ceq C1 C2 � �

A pictorial representation of two capacitors connected in parallel to a battery

A circuit diagram showing the two capacitors connected in parallel to a battery

A circuit diagram showing the equivalent capacitance of the capacitors in parallel

a b c

Two capacitors connected in parallel. All three diagrams are equivalent.

ACTIVE FIGURE 26.7

26.3 | Combinations of Capacitors 747

Series Combination

Two capacitors connected as shown in Active Figure 26.8a and the equivalent cir-cuit diagram in Active Figure 26.8b are known as a series combination of capaci-tors. The left plate of capacitor 1 and the right plate of capacitor 2 are connected to the terminals of a battery. The other two plates are connected to each other and to nothing else; hence, they form an isolated system that is initially uncharged and must continue to have zero net charge. To analyze this combination, let’s first consider the uncharged capacitors and then follow what happens immediately after a battery is connected to the circuit. When the battery is connected, electrons are transferred out of the left plate of C1 and into the right plate of C2. As this nega-tive charge accumulates on the right plate of C2, an equivalent amount of negative charge is forced off the left plate of C2, and this left plate therefore has an excess positive charge. The negative charge leaving the left plate of C2 causes negative charges to accumulate on the right plate of C1. As a result, both right plates end up with a charge 2Q and both left plates end up with a charge 1Q. Therefore, the charges on capacitors connected in series are the same:

Q 1 5 Q 2 5 Q

where Q is the charge that moved between a wire and the connected outside plate of one of the capacitors. Active Figure 26.8a shows that the total voltage DVtot across the combination is split between the two capacitors:

DVtot 5 DV1 1 DV2 (26.9)

where DV1 and DV2 are the potential differences across capacitors C1 and C2, respec-tively. In general, the total potential difference across any number of capacitors connected in series is the sum of the potential differences across the individual capacitors. Suppose the equivalent single capacitor in Active Figure 26.8c has the same effect on the circuit as the series combination when it is connected to the battery. After it is fully charged, the equivalent capacitor must have a charge of 2Q on its right plate and a charge of 1Q on its left plate. Applying the definition of capaci-tance to the circuit in Active Figure 26.8c gives

DVtot 5Q

Ceq

Substituting for the voltages in Equation 26.9, we have

Q

Ceq5 DV1 1 DV2 5

Q 1

C11

Q 2

C2

� �

C2

�V

C1�V1 �V2

�V

C1 C2

�V1 �V2�Q �Q �Q �Q

� ��V

C2Ceq C1 11 1

� �

� �

A pictorial representation of two capacitors connected in series to a battery

A circuit diagram showing the two capacitors connected in series to a battery

A circuit diagram showing the equivalent capacitance of the capacitors in series

a b c

Two capacitors connected in series. All three diagrams are equivalent.

ACTIVE FIGURE 26.8


Canceling the charges because they are all the same gives

1

Ceq5

1C1

11C2

1series combination 2

When this analysis is applied to three or more capacitors connected in series, the relationship for the equivalent capacitance is

1

C eq5

1C 1

11

C 21

1C 3

1 c 1series combination 2 (26.10)

This expression shows that (1) the inverse of the equivalent capacitance is the alge-braic sum of the inverses of the individual capacitances and (2) the equivalent capacitance of a series combination is always less than any individual capacitance in the combination.

Quick Quiz 26.3 Two capacitors are identical. They can be connected in series or in parallel. If you want the smallest equivalent capacitance for the combination, how should you connect them? (a) in series (b) in parallel (c) either way because both combinations have the same capacitance

Equivalent capacitance for capacitors in series

Example 26.3 Equivalent Capacitance

Find the equivalent capacitance between a and b for the combination of capacitors shown in Figure 26.9a. All capacitances are in microfarads.

SOLUTION

Conceptualize Study Figure 26.9a carefully and make sure you understand how the capacitors are connected.

Categorize Figure 26.9a shows that the circuit contains both series and parallel connections, so we use the rules for series and parallel combinations discussed in this section.

Analyze Using Equations 26.8 and 26.10, we reduce the combination step by step as indicated in the figure.

4.04.0

8.08.0

ba

4.0

ba

2.0

6.0 ba

4.0

8.0

ba

2.0

6.0

3.0

1.0

a b c d

Figure 26.9 (Example 26.3) To find the equivalent capacitance of the capacitors in (a), we reduce the various combinations in steps as indicated in (b), (c), and (d), using the series and parallel rules described in the text. All capacitances are in microfarads.

The 1.0-mF and 3.0-mF capacitors (upper red-brown circle in Fig. 26.9a) are in parallel. Find the equivalent capacitance from Equation 26.8:

Ceq 5 C1 1 C2 5 4.0 mF

The 2.0-mF and 6.0-mF capacitors (lower red-brown circle in Fig. 26.9a) are also in parallel:

Ceq 5 C1 1 C2 5 8.0 mF

The circuit now looks like Figure 26.9b. The two 4.0-mF capacitors (upper green circle in Fig. 26.9b) are in series. Find the equivalent capacitance from Equation 26.10:

1

C eq5

1C 1

11

C 25

14.0 mF

11

4.0 mF5

12.0 mF

C eq 5 2.0 mF

The two 8.0-mF capacitors (lower green circle in Fig. 26.9b) are also in series. Find the equivalent capacitance from Equation 26.10:

1C eq

51

C 11

1C 2

51

8.0 mF1

18.0 mF

51

4.0 mF

C eq 5 4.0 mF

The circuit now looks like Figure 26.9c. The 2.0-mF and 4.0-mF capacitors are in parallel:

Ceq 5 C1 1 C2 5 6.0 mF

26.4 | Energy Stored in a Charged Capacitor 749

26.3 cont.

26.4 Energy Stored in a Charged CapacitorBecause positive and negative charges are separated in the system of two conduc-tors in a capacitor, electric potential energy is stored in the system. Many of those who work with electronic equipment have at some time verified that a capacitor can store energy. If the plates of a charged capacitor are connected by a conductor such as a wire, charge moves between each plate and its connecting wire until the capaci-tor is uncharged. The discharge can often be observed as a visible spark. If you accidentally touch the opposite plates of a charged capacitor, your fingers act as a pathway for discharge and the result is an electric shock. The degree of shock you receive depends on the capacitance and the voltage applied to the capacitor. Such a shock could be dangerous if high voltages are present as in the power supply of a home theater system. Because the charges can be stored in a capacitor even when the system is turned off, unplugging the system does not make it safe to open the case and touch the components inside. Active Figure 26.10a shows a battery connected to a single parallel-plate capaci-tor with a switch in the circuit. Let us identify the circuit as a system. When the switch is closed (Active Fig. 26.10b), the battery establishes an electric field in the wires and charges flow between the wires and the capacitor. As that occurs, there is a transformation of energy within the system. Before the switch is closed, energy is stored as chemical potential energy in the battery. This energy is transformed dur-ing the chemical reaction that occurs within the battery when it is operating in an electric circuit. When the switch is closed, some of the chemical potential energy in the battery is converted to electric potential energy associated with the separation of positive and negative charges on the plates.

Finalize This final value is that of the single equivalent capacitor shown in Figure 26.9d. For further practice in treating circuits with combinations of capacitors, imagine a battery is connected between points a and b in Figure 26.9a so that a potential difference DV is established across the combination. Can you find the voltage across and the charge on each capacitor?

�V �V

++++++

––––––

Electricfield inwire

Electric field between plates

Chemical potentialenergy in thebattery is reduced.

Electrons move from the wire to the plate.

Electrons move from the plate to the wire, leaving the plate positively charged.

Separation of charges represents potential energy.

� ��

ES

a b

Electricfield inwire

With the switch open, the capacitor remains uncharged.

(a) A circuit consisting of a capaci-tor, a battery, and a switch. (b) When the switch is closed, the battery establishes an electric field in the wire and the capacitor becomes charged.

ACTIVE FIGURE 26.10


To calculate the energy stored in the capacitor, we shall assume a charging pro-cess that is different from the actual process described in Section 26.1 but that gives the same final result. This assumption is justified because the energy in the final configuration does not depend on the actual charge-transfer process.3 Imagine the plates are disconnected from the battery and you transfer the charge mechanically through the space between the plates as follows. You grab a small amount of posi-tive charge on the plate connected to the negative terminal and apply a force that causes this positive charge to move over to the plate connected to the positive ter-minal. Therefore, you do work on the charge as it is transferred from one plate to the other. At first, no work is required to transfer a small amount of charge dq from one plate to the other,4 but once this charge has been transferred, a small potential difference exists between the plates. Therefore, work must be done to move addi-tional charge through this potential difference. As more and more charge is trans-ferred from one plate to the other, the potential difference increases in proportion and more work is required. Suppose q is the charge on the capacitor at some instant during the charging pro-cess. At the same instant, the potential difference across the capacitor is DV 5 q/C. This relationship is graphed in Figure 26.11. From Section 25.1, we know that the work necessary to transfer an increment of charge dq from the plate carrying charge 2q to the plate carrying charge q (which is at the higher electric potential) is

dW 5 DV dq 5q

C dq

This situation is illustrated in Figure 26.11. The work required to transfer the charge dq is the area of the tan rectangle. Because 1 V 5 1 J/C, the unit for the area is the joule. The total work required to charge the capacitor from q 5 0 to some final charge q 5 Q is

W 5 3Q

0

q

C dq 5

1C

3Q

0 q dq 5

Q2

2C

The work done in charging the capacitor appears as electric potential energy U stored in the capacitor. Using Equation 26.1, we can express the potential energy stored in a charged capacitor as

U 5Q2

2C5 1

2Q DV 5 12C 1DV 22 (26.11)

Because the curve in Figure 26.11 is a straight line, the total area under the curve is that of a triangle of base Q and height DV. Equation 26.11 applies to any capacitor, regardless of its geometry. For a given capacitance, the stored energy increases as the charge and the potential difference increase. In practice, there is a limit to the maximum energy (or charge) that can be stored because, at a sufficiently large value of DV, discharge ultimately occurs between the plates. For this reason, capacitors are usually labeled with a maximum operating voltage. We can consider the energy in a capacitor to be stored in the electric field cre-ated between the plates as the capacitor is charged. This description is reason-able because the electric field is proportional to the charge on the capacitor. For a parallel-plate capacitor, the potential difference is related to the electric field through the relationship DV 5 Ed. Furthermore, its capacitance is C 5 P0A/d (Eq. 26.3). Substituting these expressions into Equation 26.11 gives

Energy stored in a charged capacitor

3This discussion is similar to that of state variables in thermodynamics. The change in a state variable such as temper-ature is independent of the path followed between the initial and final states. The potential energy of a capacitor (or any system) is also a state variable, so its change does not depend on the process followed to charge the capacitor.4We shall use lowercase q for the time-varying charge on the capacitor while it is charging to distinguish it from uppercase Q, which is the total charge on the capacitor after it is completely charged.

V

dq

qQ

�

The work required to move charge dq through the potential difference �V across the capacitor plates is given approximately by the area of the shaded rectangle.

Figure 26.11 A plot of potential dif-ference versus charge for a capacitor is a straight line having slope 1/C.

Example 26.4 Rewiring Two Charged Capacitors

Two capacitors C1 and C2 (where C1 . C2) are charged to the same initial potential difference DVi. The charged capacitors are removed from the battery, and their plates are connected with opposite polarity as in Figure 26.12a. The switches S1 and S2 are then closed as in Figure 26.12b.

(A) Find the final potential difference DVf between a and b after the switches are closed.

SOLUTION

Conceptualize Figure 26.12 helps us understand the initial and final configurations of the system. When the switches are closed, the charge on the system will redistribute between the capacitors until both capacitors have the same potential difference. Because C1 . C2, more charge exists on C1 than on C2, so the final configuration will have positive charge on the left plates as shown in Figure 26.12b.

Categorize In Figure 26.12b, it might appear as if the capacitors are connected in parallel, but there is no battery in this circuit to apply a voltage across the combination. Therefore, we cannot categorize this problem as one in which capacitors are connected in parallel. We can categorize it as a problem involving an isolated system for electric charge. The left-hand plates of the capacitors form an isolated system because they are not connected to the right-hand plates by conductors.

26.4 | Energy Stored in a Charged Capacitor 751

U 5 12 P0 A

d1E 2d 2 2 5 1

2 1P0Ad 2E 2 (26.12)

Because the volume occupied by the electric field is Ad, the energy per unit volume uE 5 U/Ad, known as the energy density, is

uE 5 12P0E 2 (26.13)

Although Equation 26.13 was derived for a parallel-plate capacitor, the expression is generally valid regardless of the source of the electric field. That is, the energy density in any electric field is proportional to the square of the magnitude of the electric field at a given point.

Quick Quiz 26.4 You have three capacitors and a battery. In which of the fol-lowing combinations of the three capacitors is the maximum possible energy stored when the combination is attached to the battery? (a) series (b) paral-lel (c) no difference because both combinations store the same amount of energy

Energy density in an electric field

Pitfall Prevention 26.4Not a New Kind of EnergyThe energy given by Equation 26.12 is not a new kind of energy. The equation describes familiar electric potential energy associated with a system of separated source charges. Equation 26.12 provides a new inter-pretation, or a new way of modeling the energy. Furthermore, Equation 26.13 correctly describes the energy density associated with any electric field, regardless of the source.

� �

Q1i�

ba

�

C1

Q 2i� �

C2

S1 S2

�

ba

�

S1 S2

Q1fC1

Q 2f C2

a b

Figure 26.12 (Example 26.4) (a) Two capacitors are charged to the same initial potential difference and con-nected together with plates of opposite sign to be in contact when the switches are closed. (b) When the switches are closed, the charges redistribute.

Analyze Write an expression for the total charge on the left-hand plates of the system before the switches are closed, noting that a negative sign for Q 2i is necessary because the charge on the left plate of capacitor C2 is negative:

(1) Q i 5 Q 1i 1 Q 2i 5 C1 DVi 2 C2 DVi 5 (C1 2 C2)DVi

After the switches are closed, the charges on the indi-vidual capacitors change to new values Q1f and Q 2f such that the potential difference is again the same across both capacitors, with a value of DVf . Write an expression for the total charge on the left-hand plates of the system after the switches are closed:

(2) Q f 5 Q 1f 1 Q 2f 5 C1 DVf 1 C2 DVf 5 (C1 1 C2)DVf

continued


26.4 cont.

One device in which capacitors have an important role is the portable defibrillator (see the chapter-opening photo on page 740). When cardiac fibrillation (random contractions) occurs, the heart produces a rapid, irregular pattern of beats. A fast discharge of energy through the heart can return the organ to its normal beat pat-tern. Emergency medical teams use portable defibrillators that contain batteries capable of charging a capacitor to a high voltage. (The circuitry actually permits the capacitor to be charged to a much higher voltage than that of the battery.) Up to 360 J is stored in the electric field of a large capacitor in a defibrillator when it is fully charged. The stored energy is released through the heart by conducting electrodes, called paddles, which are placed on both sides of the victim’s chest. The defibrillator can deliver the energy to a patient in about 2 ms (roughly equivalent to 3 000 times the power delivered to a 60-W lightbulb!). The paramedics must wait between applications of the energy because of the time interval necessary for the capacitors to become fully charged. In this application and others (e.g., cam-era flash units and lasers used for fusion experiments), capacitors serve as energy

Because the system is isolated, the initial and final charges on the system must be the same. Use this condition and Equations (1) and (2) to solve for DVf :

Q f 5 Q i S 1C 1 1 C 2 2 DVf 5 1C 1 2 C 2 2 DVi

(3) DVf 5 aC 1 2 C 2

C 1 1 C 2b DVi

(B) Find the total energy stored in the capacitors before and after the switches are closed and determine the ratio of the final energy to the initial energy.

SOLUTION

Divide Equation (5) by Equation (4) to obtain the ratio of the energies stored in the system:

Uf

Ui5

12 1C 1 2 C 2 22 1DVi 22/ 1C 1 1 C 2 2

12 1C 1 1 C 2 2 1DVi 22

(6) Uf

Ui5 aC 1 2 C 2

C 1 1 C 2b

2

Use the results of part (A) to rewrite this expres-sion in terms of DVi:

(5) Uf 5 12 1C 1 1 C 2 2 c a

C 1 2 C 2

C 1 1 C 2b DVi d

2

5 12 1C 1 2 C 2 22 1DVi 22

C 1 1 C 2

Write an expression for the total energy stored in the capacitors after the switches are closed:

Uf 5 12C 1 1DVf 22 1 1

2C 2 1DVf 22 5 12 1C 1 1 C 2 2 1DVf 22

Use Equation 26.11 to find an expression for the total energy stored in the capacitors before the switches are closed:

(4) Ui 5 12C 1 1DVi 22 1 1

2C 2 1DVi 22 5 12 1C 1 1 C 2 2 1DVi 22

Finalize The ratio of energies is less than unity, indicating that the final energy is less than the initial energy. At first, you might think the law of energy conservation has been violated, but that is not the case. The “missing” energy is transferred out of the system by the mechanism of electromagnetic waves (TER in Eq. 8.2), as we shall see in Chapter 34. Therefore, this system is isolated for electric charge, but nonisolated for energy.

WHAT IF? What if the two capacitors have the same capacitance? What would you expect to happen when the switches are closed?

Answer Because both capacitors have the same initial potential difference applied to them, the charges on the capacitors have the same magnitude. When the capacitors with opposite polarities are connected together, the equal- magnitude charges should cancel each other, leaving the capacitors uncharged. Let’s test our results to see if that is the case mathematically. In Equation (1), because the capacitances are equal, the initial charge Q i on the system of left-hand plates is zero. Equation (3) shows that DVf 5 0, which is consistent with uncharged capacitors. Finally, Equation (5) shows that Uf 5 0, which is also consistent with uncharged capacitors.

26.5 | Capacitors with Dielectrics 753

reservoirs that can be slowly charged and then quickly discharged to provide large amounts of energy in a short pulse.

26.5 Capacitors with DielectricsA dielectric is a nonconducting material such as rubber, glass, or waxed paper. We can perform the following experiment to illustrate the effect of a dielectric in a capacitor. Consider a parallel-plate capacitor that without a dielectric has a charge Q 0 and a capacitance C0. The potential difference across the capacitor is DV0 5 Q 0/C0. Figure 26.13a illustrates this situation. The potential difference is measured by a device called a voltmeter. Notice that no battery is shown in the figure; also, we must assume no charge can flow through an ideal voltmeter. Hence, there is no path by which charge can flow and alter the charge on the capacitor. If a dielectric is now inserted between the plates as in Figure 26.13b, the voltmeter indicates that the voltage between the plates decreases to a value DV. The voltages with and without the dielectric are related by a factor k as follows:

DV 5DV0

k

Because DV , DV0, we see that k . 1. The dimensionless factor k is called the dielec-tric constant of the material. The dielectric constant varies from one material to another. In this section, we analyze this change in capacitance in terms of electrical parameters such as electric charge, electric field, and potential difference; Section 26.7 describes the microscopic origin of these changes. Because the charge Q 0 on the capacitor does not change, the capacitance must change to the value

C 5Q 0

DV5

Q 0

DV0/k5 k

Q 0

DV0

C 5 kC 0 (26.14)

That is, the capacitance increases by the factor k when the dielectric completely fills the region between the plates.5 Because C0 5 P0A/d (Eq. 26.3) for a parallel-plate

Capacitance of a capacitor filled with a material of dielectric constant k

Pitfall Prevention 26.5Is the Capacitor Connected to a Battery?For problems in which a capaci-tor is modified (by insertion of a dielectric, for example), you must note whether modifications to the capacitor are being made while the capacitor is connected to a battery or after it is disconnected. If the capaci-tor remains connected to the bat-tery, the voltage across the capacitor necessarily remains the same. If you disconnect the capacitor from the battery before making any modifica-tions to the capacitor, the capacitor is an isolated system and its charge remains the same.

C0 Q 0

��

C Q 0

Dielectric

VV0

��

� �

The potential difference across the charged capacitor is initially �V0.

After the dielectric is inserted between the plates, the charge remains the same, but the potential difference decreases and the capacitance increases.

a b

Figure 26.13 A charged capacitor (a) before and (b) after insertion of a dielectric between the plates.

5 If the dielectric is introduced while the potential difference is held constant by a battery, the charge increases to a value Q 5 kQ 0. The additional charge comes from the wires attached to the capacitor, and the capacitance again increases by the factor k.


capacitor, we can express the capacitance of a parallel-plate capacitor filled with a dielectric as

C 5 k P0Ad

(26.15)

From Equations 26.3 and 26.15, it would appear that the capacitance could be made very large by inserting a dielectric between the plates and decreasing d. In practice, the lowest value of d is limited by the electric discharge that could occur through the dielectric medium separating the plates. For any given separation d, the maximum voltage that can be applied to a capacitor without causing a discharge depends on the dielectric strength (maximum electric field) of the dielectric. If the magnitude of the electric field in the dielectric exceeds the dielectric strength, the insulating properties break down and the dielectric begins to conduct. Physical capacitors have a specification called by a variety of names, including working voltage, breakdown voltage, and rated voltage. This parameter represents the largest voltage that can be applied to the capacitor without exceeding the dielectric strength of the dielectric material in the capacitor. Consequently, when selecting a capacitor for a given application, you must consider its capacitance as well as the expected voltage across the capacitor in the circuit, making sure the expected volt-age is smaller than the rated voltage of the capacitor. Insulating materials have values of k greater than unity and dielectric strengths greater than that of air as Table 26.1 indicates. Therefore, a dielectric provides the following advantages:

• An increase in capacitance• An increase in maximum operating voltage• Possible mechanical support between the plates, which allows the plates to be

close together without touching, thereby decreasing d and increasing C

Types of Capacitors

Many capacitors are built into integrated circuit chips, but some electrical devices still use stand-alone capacitors. Commercial capacitors are often made from metal-lic foil interlaced with thin sheets of either paraffin-impregnated paper or Mylar

Approximate Dielectric Constants and Dielectric Strengths of Various Materials at Room Temperature

Material Dielectric Constant k Dielectric Strengtha (106 V/m)

Air (dry) 1.000 59 3Bakelite 4.9 24Fused quartz 3.78 8Mylar 3.2 7Neoprene rubber 6.7 12Nylon 3.4 14Paper 3.7 16Paraffin-impregnated paper 3.5 11Polystyrene 2.56 24Polyvinyl chloride 3.4 40Porcelain 6 12Pyrex glass 5.6 14Silicone oil 2.5 15Strontium titanate 233 8Teflon 2.1 60Vacuum 1.000 00 —Water 80 —

aThe dielectric strength equals the maximum electric field that can exist in a dielectric without electrical breakdown. These values depend strongly on the presence of impurities and flaws in the materials.

TABLE 26.1

26.5 | Capacitors with Dielectrics 755

as the dielectric material. These alternate layers of metallic foil and dielectric are rolled into a cylinder to form a small package (Fig. 26.14a). High-voltage capacitors commonly consist of a number of interwoven metallic plates immersed in silicone oil (Fig. 26.14b). Small capacitors are often constructed from ceramic materials. Often, an electrolytic capacitor is used to store large amounts of charge at relatively low voltages. This device, shown in Figure 26.14c, consists of a metallic foil in con-tact with an electrolyte, a solution that conducts electricity by virtue of the motion of ions contained in the solution. When a voltage is applied between the foil and the electrolyte, a thin layer of metal oxide (an insulator) is formed on the foil, and this layer serves as the dielectric. Very large values of capacitance can be obtained in an electrolytic capacitor because the dielectric layer is very thin and therefore the plate separation is very small. Electrolytic capacitors are not reversible as are many other capacitors. They have a polarity, which is indicated by positive and negative signs marked on the device. When electrolytic capacitors are used in circuits, the polarity must be correct. If the polarity of the applied voltage is the opposite of what is intended, the oxide layer is removed and the capacitor conducts electricity instead of storing charge. Variable capacitors (typically 10 to 500 pF) usually consist of two interwoven sets of metallic plates, one fixed and the other movable, and contain air as the dielec-tric (Fig. 26.15). These types of capacitors are often used in radio tuning circuits.

Quick Quiz 26.5 If you have ever tried to hang a picture or a mirror, you know it can be difficult to locate a wooden stud in which to anchor your nail or screw. A carpenter’s stud finder is a capacitor with its plates arranged side by side instead of facing each other as shown in Figure 26.16. When the device is moved over a stud, does the capacitance (a) increase or (b) decrease?

Plates

ElectrolyteCase

Metallic foil � oxide layer

Contacts

Metal foil

Paper

An electrolytic capacitor

Oil

a b c

A tubular capacitor whose plates are separated by paper and then rolled into a cylinder

A high-voltage capacitor consisting of many parallel plates separated by insulating oil

Figure 26.14 Three commercial capacitor designs.

When one set of metal plates is rotated so as to lie between a fixed set of plates, the capacitance of the device changes.

Figure 26.15 A variable capacitor.

. C

enga

ge L

earn

ing/

Geor

ge S

empl

e

Capacitorplates

Stud finder

Wallboard

Stud

a b

The materials between the plates of the capacitor are the wallboard and air.

When the capacitor moves across a stud in the wall, the materials between the plates are the wallboard and the wood stud. The change in the dielectric constant causes a signal light to illuminate.

Figure 26.16 (Quick Quiz 26.5) A stud finder.

Example 26.5 Energy Stored Before and After

A parallel-plate capacitor is charged with a battery to a charge Q 0. The battery is then removed, and a slab of material that has a dielectric constant k is inserted between the plates. Identify the system as the capacitor and the dielectric. Find the energy stored in the system before and after the dielectric is inserted.

SOLUTION

Conceptualize Think about what happens when the dielectric is inserted between the plates. Because the battery has been removed, the charge on the capacitor must remain the same. We know from our earlier discussion, however, that the capacitance must change. Therefore, we expect a change in the energy of the system. continued


26.5 cont.

Categorize Because we expect the energy of the system to change, we model it as a nonisolated system involving a capaci-tor and a dielectric.

Use Equation 26.14 to replace the capacitance C : U 5Q 0

2

2kC05

U0

k

Find the energy stored in the capacitor after the dielec-tric is inserted between the plates:

U 5Q 0

2

2C

Analyze From Equation 26.11, find the energy stored in the absence of the dielectric:

U0 5Q 0

2

2C0

Finalize Because k . 1, the final energy is less than the initial energy. We can account for the decrease in energy of the system by performing an experiment and noting that the dielectric, when inserted, is pulled into the device. To keep the dielectric from accelerating, an external agent must do negative work (W in Eq. 8.2) on the dielectric, which is simply the difference U 2 U0.

26.6 Electric Dipole in an Electric FieldWe have discussed the effect on the capacitance of placing a dielectric between the plates of a capacitor. In Section 26.7, we shall describe the microscopic origin of this effect. Before we can do so, however, let’s expand the discussion of the electric dipole introduced in Section 23.4 (see Example 23.5). The electric dipole consists of two charges of equal magnitude and opposite sign separated by a distance 2a as shown in Figure 26.17. The electric dipole moment of this configuration is defined as the vector pS directed from 2q toward 1q along the line joining the charges and having magnitude

p ; 2aq (26.16)

Now suppose an electric dipole is placed in a uniform electric field ES

and makes an angle u with the field as shown in Figure 26.18. We identify E

S as the field external

to the dipole, established by some other charge distribution, to distinguish it from the field due to the dipole, which we discussed in Section 23.4. The electric forces acting on the two charges are equal in magnitude (F 5 qE) and opposite in direction as shown in Figure 26.18. Therefore, the net force on the dipole is zero. The two forces produce a net torque on the dipole, however; the dipole is a rigid object under a net torque. As a result, the dipole rotates in the direction that brings the dipole moment vector into greater alignment with the field. The torque due to the force on the positive charge about an axis through O in Figure 26.18 has magnitude Fa sin u, where a sin u is the moment arm of F about O. This force tends to produce a clockwise rotation. The torque about O on the nega-tive charge is also of magnitude Fa sin u; here again, the force tends to produce a clockwise rotation. Therefore, the magnitude of the net torque about O is

t 5 2Fa sin u

Because F 5 qE and p 5 2aq, we can express t as

t 5 2aqE sin u 5 pE sin u (26.17)

Based on this expression, it is convenient to express the torque in vector form as the cross product of the vectors pS and E

S:

tS

5 pS 3 ES

(26.18)Torque on an electric dipole in an external electric field

�q

�q

2a

pS �

�

The electric dipole moment p is directed from �q toward �q.

S

Figure 26.17 An electric dipole consists of two charges of equal mag-nitude and opposite sign separated by a distance of 2a.

� q

�q

O

�

�

u

ES

�FS

FS

pS

The dipole moment p is at an angle u to the field, causing the dipole to experience a torque.

S

Figure 26.18 An electric dipole in a uniform external electric field.

26.6 | Electric Dipole in an Electric Field 757

We can also model the system of the dipole and the external electric field as an isolated system for energy. Let’s determine the potential energy of the system as a function of the dipole’s orientation with respect to the field. To do so, recognize that work must be done by an external agent to rotate the dipole through an angle so as to cause the dipole moment vector to become less aligned with the field. The work done is then stored as potential energy in the system. The work dW required to rotate the dipole through an angle du is dW 5 t du (see Eq. 10.22). Because t 5 pE sin u and the work results in an increase in the potential energy U, we find that for a rotation from ui to uf , the change in potential energy of the system is

Uf 2 Ui 5 3uf

ui

t du 5 3uf

ui

pE sin u du 5 pE 3uf

ui

sin u du

5 pE 32cos u 4u fui

5 pE 1cos ui 2 cos uf 2 The term that contains cos ui is a constant that depends on the initial orientation of the dipole. It is convenient to choose a reference angle of ui 5 908 so that cos ui 5 cos 908 5 0. Furthermore, let’s choose Ui 5 0 at ui 5 908 as our reference value of potential energy. Hence, we can express a general value of U 5 Uf as

U 5 2pE cos u (26.19)

We can write this expression for the potential energy of a dipole in an electric field as the dot product of the vectors pS and E

S:

U 5 2pS ? ES

(26.20)

To develop a conceptual understanding of Equation 26.19, compare it with the expression for the potential energy of the system of an object in the Earth’s gravita-tional field, U 5 mgy (see Chapter 7). This expression includes a parameter associ-ated with the object placed in the gravitational field, its mass m. Likewise, Equation 26.19 includes a parameter of the object in the electric field, its dipole moment p. The gravitational expression includes the magnitude of the gravitational field g. Similarly, Equation 26.19 includes the magnitude of the electric field E. So far, these two contributions to the potential energy expressions appear analogous. The final contribution, however, is somewhat different in the two cases. In the gravi-tational expression, the potential energy depends on the vertical position of the object, measured by y. In Equation 26.19, the potential energy depends on the angle u through which the dipole is rotated. In both cases, the configuration of the system is being changed. In the gravitational case, the change involves moving an object in a translational sense, whereas in the electrical case, the change involves moving an object in a rotational sense. In both cases, however, once the change is made, the system tends to return to the original configuration when the object is released: the object of mass m falls toward the ground, and the dipole begins to rotate back toward the configuration in which it is aligned with the field. There-fore, apart from the type of motion, the expressions for potential energy in these two cases are similar. Molecules are said to be polarized when a separation exists between the average position of the negative charges and the average position of the positive charges in the molecule. In some molecules such as water, this condition is always present; such molecules are called polar molecules. Molecules that do not possess a perma-nent polarization are called nonpolar molecules. We can understand the permanent polarization of water by inspecting the geom-etry of the water molecule. The oxygen atom in the water molecule is bonded to the hydrogen atoms such that an angle of 1058 is formed between the two bonds (Fig. 26.19). The center of the negative charge distribution is near the oxygen atom, and the center of the positive charge distribution lies at a point midway along the line joining the hydrogen atoms (the point labeled 3 in Fig. 26.19). We can model the water molecule and other polar molecules as dipoles because the average positions

Potential energy of the system of an electric dipole in an external electric field

O

HH 105�

�

� �

The center of the positive charge distribution is at the point .

Figure 26.19 The water molecule, H2O, has a permanent polariza-tion resulting from its nonlinear geometry.


of the positive and negative charges act as point charges. As a result, we can apply our discussion of dipoles to the behavior of polar molecules. Washing with soap and water is a household scenario in which the dipole struc-ture of water is exploited. Grease and oil are made up of nonpolar molecules, which are generally not attracted to water. Plain water is not very useful for removing this type of grime. Soap contains long molecules called surfactants. In a long molecule, the polarity characteristics of one end of the molecule can be different from those at the other end. In a surfactant molecule, one end acts like a nonpolar molecule and the other acts like a polar molecule. The nonpolar end can attach to a grease or oil molecule, and the polar end can attach to a water molecule. Therefore, the soap serves as a chain, linking the dirt and water molecules together. When the water is rinsed away, the grease and oil go with it. A symmetric molecule (Fig. 26.20a) has no permanent polarization, but polar-ization can be induced by placing the molecule in an electric field. A field directed to the left as in Figure 26.20b causes the center of the negative charge distribution to shift to the right relative to the positive charges. This induced polarization is the effect that predominates in most materials used as dielectrics in capacitors.

��

� �

�

�

ES

a

b

Figure 26.20 (a) A linear sym-metric molecule has no permanent polarization. (b) An external electric field induces a polarization in the molecule.

Example 26.6 The H2O Molecule

The water (H2O) molecule has an electric dipole moment of 6.3 3 10230 C ? m. A sample contains 1021 water molecules, with the dipole moments all oriented in the direction of an electric field of magnitude 2.5 3 105 N/C. How much work is required to rotate the dipoles from this orientation (u 5 08) to one in which all the moments are perpendicular to the field (u 5 908)?

SOLUTION

Conceptualize When all the dipoles are aligned with the electric field, the dipoles–electric field system has the mini-mum potential energy. This energy has a negative value given by the product of the right side of Equation 26.19, evalu-ated at 08, and the number N of dipoles. Work must be done to rotate all the dipoles of the system by 908 because the system’s potential energy is raised to a higher value of zero.

Categorize We use Equation 26.19 to evaluate the potential energy, so we categorize this example as a substitution problem.

Write the appropriate reduction of the conservation of energy equation, Equation 8.2, for this situation:

(1) DU 5 W

Use Equation 26.19 to evaluate the initial and final potential energies of the system and Equation (1) to cal-culate the work required to rotate the dipoles:

W 5 U908 2 U08 5 (2NpE cos 908) 2 (2NpE cos 08)

5 NpE 5 (1021)(6.3 3 10230 C ? m)(2.5 3 105 N/C)

5 1.6 3 1023 J

26.7 An Atomic Description of DielectricsIn Section 26.5, we found that the potential difference DV0 between the plates of a capacitor is reduced to DV0/k when a dielectric is introduced. The potential differ-ence is reduced because the magnitude of the electric field decreases between the plates. In particular, if E

S0 is the electric field without the dielectric, the field in the

presence of a dielectric is

ES

5ES

0

k (26.21)

First consider a dielectric made up of polar molecules placed in the electric field between the plates of a capacitor. The dipoles (that is, the polar molecules making up the dielectric) are randomly oriented in the absence of an electric field as shown in Figure 26.21a. When an external field E

S0 due to charges on the capacitor plates

26.7 | An Atomic Description of Dielectrics 759

is applied, a torque is exerted on the dipoles, causing them to partially align with the field as shown in Figure 26.21b. The dielectric is now polarized. The degree of alignment of the molecules with the electric field depends on temperature and the magnitude of the field. In general, the alignment increases with decreasing tem-perature and with increasing electric field. If the molecules of the dielectric are nonpolar, the electric field due to the plates produces an induced polarization in the molecule. These induced dipole moments tend to align with the external field, and the dielectric is polarized. Therefore, a dielectric can be polarized by an external field regardless of whether the molecules in the dielectric are polar or nonpolar. With these ideas in mind, consider a slab of dielectric material placed between the plates of a capacitor so that it is in a uniform electric field E

S0 as shown in Figure

26.21b. The electric field due to the plates is directed to the right and polarizes the dielectric. The net effect on the dielectric is the formation of an induced positive surface charge density sind on the right face and an equal-magnitude negative sur-face charge density 2sind on the left face as shown in Figure 26.21c. Because we can model these surface charge distributions as being due to charged parallel plates, the induced surface charges on the dielectric give rise to an induced electric field ES

ind in the direction opposite the external field ES

0. Therefore, the net electric field ES

in the dielectric has a magnitude

E 5 E0 2 E ind (26.22)

In the parallel-plate capacitor shown in Figure 26.22, the external field E0 is related to the charge density s on the plates through the relationship E0 5 s/P0. The induced electric field in the dielectric is related to the induced charge density sind through the relationship Eind 5 sind/P0. Because E 5 E0/k 5 s/kP0, substitution into Equation 26.22 gives

s

kP05

s

P02

sind

P0

sind 5 ak 2 1k

bs (26.23)

Because k . 1, this expression shows that the charge density sind induced on the dielectric is less than the charge density s on the plates. For instance, if k 5 3, the induced charge density is two-thirds the charge density on the plates. If no dielec-tric is present, then k 5 1 and sind 5 0 as expected. If the dielectric is replaced by an electrical conductor for which E 5 0, however, Equation 26.22 indicates that E0 5 Eind, which corresponds to sind 5 s. That is, the surface charge induced on the conductor is equal in magnitude but opposite in sign to that on the plates, resulting in a net electric field of zero in the conductor (see Fig. 24.14).

�

�

�

�

�

�

�

�

�

�

�sind inds

�

�

�

�

�

�

�

�

�

�

�

�

�

�

�

�

�

�

�

�

�

�

�

�

E0S

E0S

EindS

a b c

��

��

��

��

��

��

��

��

��

��

��

��

��

��

� � � ��

��

��

��

��

Polar molecules are randomly oriented in the absence of an external electric field.

When an external electric field is applied, the molecules partially align with the field.

The charged edges of the dielectric can be modeled as an additional pair of parallel plates establishing an electric field Eind in the direction opposite that of E0.

S

S

Figure 26.21 (a) Polar molecules in a dielectric. (b) An electric field is applied to the dielectric. (c) Details of the electric field inside the dielectric.

�sind �ss sind

�

�

�

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The induced charge density sind on the dielectric is less than the charge density s on the plates.

Figure 26.22 Induced charge on a dielectric placed between the plates of a charged capacitor.


Example 26.7 Effect of a Metallic Slab

A parallel-plate capacitor has a plate separation d and plate area A. An uncharged metallic slab of thickness a is inserted midway between the plates.

(A) Find the capacitance of the device.

SOLUTION

Conceptualize Figure 26.23a shows the metallic slab between the plates of the capacitor. Any charge that appears on one plate of the capacitor must induce a charge of equal magni-tude and opposite sign on the near side of the slab as shown in Figure 26.23a. Consequently, the net charge on the slab remains zero and the electric field inside the slab is zero.

Categorize The planes of charge on the metallic slab’s upper and lower edges are identical to the distribution of charges on the plates of a capacitor. The metal between the slab’s edges serves only to make an electrical connection between the edges. Therefore, we can model the edges of the slab as conducting planes and the bulk of the slab as a wire. As a result, the capacitor in Figure 26.23a is equivalent to two capacitors in series, each having a plate separation (d 2 a)/2 as shown in Figure 26.23b.

d a

(d � a)/2

s

s � � � � ��

� � � � �

� � � � �

� � � � �

� � � � �

� � � � ��

�s

�s

a b

(d � a)/2

(d � a)/2

(d � a)/2

Figure 26.23 (Example 26.7) (a) A parallel-plate capaci-tor of plate separation d partially filled with a metallic slab of thickness a. (b) The equivalent circuit of the device in (a) consists of two capacitors in series, each having a plate separation (d 2 a)/2.

Analyze Use Equation 26.3 and the rule for adding two capacitors in series (Eq. 26.10) to find the equivalent capacitance in Figure 26.23b:

1C

51

C 1 1

1C 2

51P0A

1d 2 a 2/2

11P0A

1d 2 a 2/2

C 5 P0A

d 2 a

(B) Show that the capacitance of the original capacitor is unaffected by the insertion of the metallic slab if the slab is infinitesimally thin.

SOLUTION

In the result for part (A), let a S 0: C 5 limaS0

a P0Ad 2 a

b 5P0Ad

Finalize The result of part (B) is the original capacitance before the slab is inserted, which tells us that we can insert an infinitesimally thin metallic sheet between the plates of a capacitor without affecting the capacitance. We use this fact in the next example.

WHAT IF? What if the metallic slab in part (A) is not midway between the plates? How would that affect the capacitance?

Answer Let’s imagine moving the slab in Figure 26.23a upward so that the distance between the upper edge of the slab and the upper plate is b. Then, the distance between the lower edge of the slab and the lower plate is d 2 b 2 a. As in part (A), we find the total capacitance of the series combination:

1C

51C1

11C2

51

P0A/b1

1P0A/ 1d 2 b 2 a 2

5bP0A

1d 2 b 2 a

P0A5

d 2 aP0A

S C 5P0A

d 2 a

which is the same result as found in part (A). The capacitance is independent of the value of b, so it does not matter where the slab is located. In Figure 26.23b, when the central structure is moved up or down, the decrease in plate separa-tion of one capacitor is compensated by the increase in plate separation for the other.

| Summary 761

Example 26.8 A Partially Filled Capacitor

A parallel-plate capacitor with a plate separation d has a capacitance C0 in the absence of a dielectric. What is the capacitance when a slab of dielectric material of dielectric constant k and thickness fd is inserted between the plates (Fig. 26.24a), where f is a fraction between 0 and 1?

SOLUTIONConceptualize In our previous discussions of dielectrics between the plates of a capacitor, the dielectric filled the volume between the plates. In this example, only part of the volume between the plates contains the dielectric material.Categorize In Example 26.7, we found that an infinitesi-mally thin metallic sheet inserted between the plates of a capacitor does not affect the capacitance. Imagine sliding an infinitesimally thin metallic slab along the bottom face of the dielectric shown in Figure 26.24a. We can model this system as a series combination of two capacitors as shown in Figure 26.24b. One capacitor has a plate separation fdand is filled with a dielectric; the other has a plate separa-tion (1 2 f )d and has air between its plates.

fd

(1 � f )dd

C 1

C 2(1 � f )d

k

k

a b

fd

Figure 26.24 (Example 26.8) (a) A parallel-plate capacitor of plate separation d partially filled with a dielectric of thick-ness fd. (b) The equivalent circuit of the capacitor consists of two capacitors connected in series.

Invert and substitute for the capacitance without the dielectric, C0 5 P0A/d :

C 5k

f 1 k 11 2 f 2 P0Ad

5k

f 1 k 11 2 f 2 C 0

Find the equivalent capacitance C from Equation 26.10 for two capacitors combined in series:

1C

51

C 11

1C 2

5fd

kP0A111 2 f 2dP0 A

1C

5fd

kP0A1

k 11 2 f 2dkP0A

5f 1 k 11 2 f 2

k

dP0A

Analyze Evaluate the two capacitances in Figure 26.24b from Equation 26.15:

C 1 5kP0A

fd and C 2 5

P0A11 2 f 2d

Finalize Let’s test this result for some known limits. If f S 0, the dielectric should disappear. In this limit, C S C0, which is consistent with a capacitor with air between the plates. If f S 1, the dielectric fills the volume between the plates. In this limit, C S kC0, which is consistent with Equation 26.14.

SummaryDefinitions

A capacitor consists of two conductors carrying charges of equal mag-nitude and opposite sign. The capacitance C of any capacitor is the ratio of the charge Q on either conductor to the potential difference DV between them:

C ;Q

DV (26.1)

The capacitance depends only on the geometry of the conductors and not on an external source of charge or potential difference. The SI unit of capacitance is coulombs per volt, or the farad (F): 1 F 5 1 C/V.

The electric dipole moment pS of an elec-tric dipole has a magnitude

p ; 2aq (26.16)

where 2a is the distance between the charges q and 2q. The direction of the electric dipole moment vector is from the negative charge toward the positive charge.

continued


capacitor (d) slightly less than the capacitance of the small capacitor

5. If three unequal capacitors, initially uncharged, are connected in series across a battery, which of the follow-ing statements is true? (a) The equivalent capacitance is greater than any of the individual capacitances. (b) The largest voltage appears across the smallest capacitance. (c) The largest voltage appears across the largest capaci-tance. (d) The capacitor with the largest capacitance has the greatest charge. (e) The capacitor with the smallest capacitance has the smallest charge.

6. (i) Rank the following five capacitors from greatest to small-est capacitance, noting any cases of equality. (a) a 20-mF capacitor with a 4-V potential difference between its plates (b) a 30-mF capacitor with charges of magnitude 90 mC on each plate (c) a capacitor with charges of magnitude 80 mC on its plates, differing by 2 V in potential, (d) a 10-mF capac-itor storing energy 125 mJ (e) a capacitor storing energy 250 mJ with a 10-V potential difference (ii) Rank the same capacitors in part (i) from largest to smallest according to

1. True or False? (a) From the definition of capacitance C 5Q /DV, it follows that an uncharged capacitor has a capaci-tance of zero. (b) As described by the definition of capaci-tance, the potential difference across an uncharged capacitor is zero.

2. By what factor is the capacitance of a metal sphere multi-plied if its volume is tripled? (a) 3 (b) 31/3 (c) 1 (d) 321/3

(e) 13

3. An electronics technician wishes to construct a parallel-plate capacitor using rutile (k 5 100) as the dielectric. The area of the plates is 1.00 cm2. What is the capacitance if the rutile thickness is 1.00 mm? (a) 88.5 pF (b) 177 pF (c) 8.85 mF (d) 100 mF (e) 35.4 mF

4. A capacitor with very large capacitance is in series with another capacitor with very small capacitance. What is the equivalent capacitance of the combination? (a) slightly greater than the capacitance of the large capacitor (b) slightly less than the capacitance of the large capaci-tor (c) slightly greater than the capacitance of the small

Concepts and Principles

If two or more capacitors are connected in parallel, the potential difference is the same across all capacitors. The equivalent capaci-tance of a parallel combination of capacitors is

C eq 5 C 1 1 C 2 1 C 3 1 . . . (26.8)

If two or more capacitors are connected in series, the charge is the same on all capacitors, and the equivalent capacitance of the series combination is given by

1C eq

51

C 11

1C 2

11

C 31 c (26.10)

These two equations enable you to simplify many electric circuits by replacing multiple capacitors with a single equivalent capacitance.

Energy is stored in a charged capacitor because the charging process is equivalent to the transfer of charges from one conduc-tor at a lower electric potential to another conductor at a higher potential. The energy stored in a capacitor of capacitance C with charge Q and potential difference DV is

U 5Q2

2C5 1

2Q DV 5 12C 1DV 22 (26.11)

When a dielectric material is inserted between the plates of a capacitor, the capacitance increases by a dimensionless factor k, called the dielectric constant:

C 5 kC 0 (26.14)

where C 0 is the capacitance in the absence of the dielectric.

The torque acting on an electric dipole in a uniform electric field E

S is

tS

5 pS 3 ES

(26.18)

The potential energy of the system of an electric dipole in a uniform external electric field E

S is

U 5 2pS ? ES

(26.20)

Objective Questions denotes answer available in Student Solutions Manual/Study Guide

| Conceptual Questions 763

differences across the capacitors are the same only if the capacitances are the same. (ii) The capacitors are recon-nected in series, and the combination is again connected to the battery. From the same choices, choose the one that is true.

11. A parallel-plate capacitor filled with air carries a charge Q. The battery is disconnected, and a slab of material with dielectric constant k 5 2 is inserted between the plates. Which of the following statements is true? (a) The voltage across the capacitor decreases by a factor of 2. (b) The volt-age across the capacitor is doubled. (c) The charge on the plates is doubled. (d) The charge on the plates decreases by a factor of 2. (e) The electric field is doubled.

12. Assume a device is designed to obtain a large potential dif-ference by first charging a bank of capacitors connected in parallel and then activating a switch arrangement that in effect disconnects the capacitors from the charging source and from each other and reconnects them all in a series arrangement. The group of charged capacitors is then discharged in series. What is the maximum potential dif-ference that can be obtained in this manner by using ten 500-mF capacitors and an 800-V charging source? (a) 500 V (b) 8.00 kV (c) 400 kV (d) 800 V (e) 0

13. A fully charged parallel-plate capacitor remains connected to a battery while you slide a dielectric between the plates. Do the following quantities (a) increase, (b) decrease, or (c) stay the same? (i) C (ii) Q (iii) DV (iv) the energy stored in the capacitor

14. You charge a parallel-plate capacitor, remove it from the battery, and prevent the wires connected to the plates from touching each other. When you increase the plate separa-tion, do the following quantities (a) increase, (b) decrease, or (c) stay the same? (i) C (ii) Q (iii) E between the plates (iv) DV

the potential difference between the plates. (iii) Rank the capacitors in part (i) in the order of the magnitudes of the charges on their plates. (iv) Rank the capacitors in part (i) in the order of the energy they store.

7. (i) What happens to the magnitude of the charge on each plate of a capacitor if the potential difference between the conductors is doubled? (a) It becomes four times larger. (b) It becomes two times larger. (c) It is unchanged. (d) It becomes one-half as large. (e) It becomes one-fourth as large. (ii) If the potential difference across a capacitor is doubled, what happens to the energy stored? Choose from the same possibilities as in part (i).

8. A parallel-plate capacitor is connected to a battery. What happens to the stored energy if the plate separation is dou-bled while the capacitor remains connected to the battery? (a) It remains the same. (b) It is doubled. (c) It decreases by a factor of 2. (d) It decreases by a factor of 4. (e) It increases by a factor of 4.

9. A parallel-plate capacitor is charged and then is discon-nected from the battery. By what factor does the stored energy change when the plate separation is then doubled? (a) It becomes four times larger. (b) It becomes two times larger. (c) It stays the same. (d) It becomes one-half as large. (e) It becomes one-fourth as large.

10. (i) A battery is attached to several different capacitors connected in parallel. Which of the following statements is true? (a) All capacitors have the same charge, and the equivalent capacitance is greater than the capacitance of any of the capacitors in the group. (b) The capacitor with the largest capacitance carries the smallest charge. (c) The potential difference across each capacitor is the same, and the equivalent capacitance is greater than any of the capacitors in the group. (d) The capacitor with the small-est capacitance carries the largest charge. (e) The potential

Conceptual Questions denotes answer available in Student Solutions Manual/Study Guide

1. The sum of the charges on both plates of a capacitor is zero. What does a capacitor store?

2. Assume you want to increase the maximum operating volt-age of a parallel-plate capacitor. Describe how you can do that with a fixed plate separation.

3. If you were asked to design a capacitor in which small size and large capacitance were required, what would be the two most important factors in your design?

4. Because the charges on the plates of a parallel-plate capac-itor are opposite in sign, they attract each other. Hence, it would take positive work to increase the plate separa-tion. What type of energy in the system changes due to the external work done in this process?

5. Explain why the work needed to move a particle with charge Q through a potential difference DV is W 5 Q DV, whereas

the energy stored in a charged capacitor is U 5 12Q DV.

Where does the factor 12 come from?

6. An air-filled capacitor is charged, then disconnected from the power supply, and finally connected to a voltmeter. Explain how and why the potential difference changes when a dielectric is inserted between the plates of the capacitor.

7. (a) Why is it dangerous to touch the terminals of a high-voltage capacitor even after the voltage source that charged the capacitor is disconnected from the capacitor? (b) What can be done to make the capacitor safe to handle after the voltage source has been removed?

8. Explain why a dielectric increases the maximum operating voltage of a capacitor even though the physical size of the capacitor doesn’t change.


8. An air-filled spherical capacitor is constructed with inner- and outer-shell radii of 7.00 cm and 14.0 cm, respec-tively. (a) Calculate the capacitance of the device. (b) What potential difference between the spheres results in a 4.00-mC charge on the capacitor?

9. An isolated, charged conducting sphere of radius 12.0 cm creates an electric field of 4.90 3 104 N/C at a distance 21.0 cm from its center. (a) What is its surface charge den-sity? (b) What is its capacitance?

10. A variable air capacitor used in a radio tuning cir-cuit is made of N semicircu-lar plates, each of radius R and positioned a distance d from its neighbors, to which it is electrically connected. As shown in Figure P26.10, a second identical set of plates is enmeshed with the first set. Each plate in the second set is halfway between two plates of the first set. The second set can rotate as a unit. Determine the capacitance as a function of the angle of rotation u, where u 5 0 corresponds to the maximum capacitance.

11. An air-filled capacitor consists of two parallel plates, each with an area of 7.60 cm2, separated by a distance of 1.80 mm. A 20.0-V potential difference is applied to these plates. Calculate (a) the electric field between the plates, (b) the surface charge density, (c) the capacitance, and (d) the charge on each plate.

12. Review. A small object of mass m carries a charge q and is suspended by a thread between the vertical plates of a parallel-plate capacitor. The plate separation is d. If the thread makes an angle u with the vertical, what is the potential difference between the plates?

Section 26.3 Combinations of Capacitors

13. Two capacitors, C1 5 5.00 mF and C2 5 12.0 mF, are con-nected in parallel, and the resulting combination is con-nected to a 9.00-V battery. Find (a) the equivalent capaci-tance of the combination, (b) the potential difference across each capacitor, and (c) the charge stored on each capacitor.

14. What If? The two capacitors of Problem 13 (C1 5 5.00 mF and C2 5 12.0 mF) are now connected in series and to a 9.00-V battery. Find (a) the equivalent capacitance of the

Section 26.1 Definition of Capacitance

1. (a) How much charge is on each plate of a 4.00-mF capaci-tor when it is connected to a 12.0-V battery? (b) If this same capacitor is connected to a 1.50-V battery, what charge is stored?

2. Two conductors having net charges of 110.0 mC and 210.0 mC have a potential difference of 10.0 V between them. (a) Determine the capacitance of the system. (b) What is the potential difference between the two con-ductors if the charges on each are increased to 1100 mC and 2100 mC?

3. (a) When a battery is connected to the plates of a 3.00-mF capacitor, it stores a charge of 27.0 mC. What is the volt-age of the battery? (b) If the same capacitor is connected to another battery and 36.0 mC of charge is stored on the capacitor, what is the voltage of the battery?

Section 26.2 Calculating Capacitance

4. An air-filled parallel-plate capacitor has plates of area 2.30 cm2 separated by 1.50 mm. (a) Find the value of its capacitance. The capacitor is connected to a 12.0-V bat-tery. (b) What is the charge on the capacitor? (c) What is the magnitude of the uniform electric field between the plates?

5. A 50.0-m length of coaxial cable has an inner conduc-tor that has a diameter of 2.58 mm and carries a charge of 8.10 mC. The surrounding conductor has an inner diameter of 7.27 mm and a charge of 28.10 mC. Assume the region between the conductors is air. (a) What is the capacitance of this cable? (b) What is the potential difference between the two conductors?

6. (a) Regarding the Earth and a cloud layer 800 m above the Earth as the “plates” of a capacitor, calculate the capaci-tance of the Earth–cloud layer system. Assume the cloud layer has an area of 1.00 km2 and the air between the cloud and the ground is pure and dry. Assume charge builds up on the cloud and on the ground until a uniform electric field of 3.00 3 106 N/C throughout the space between them makes the air break down and conduct electricity as a lightning bolt. (b) What is the maximum charge the cloud can hold?

7. When a potential difference of 150 V is applied to the plates of a parallel-plate capacitor, the plates carry a sur-face charge density of 30.0 nC/cm2. What is the spacing between the plates?

Problems

denotes asking for quantitative and conceptual reasoning

denotes symbolic reasoning problem

denotes Master It tutorial available in Enhanced WebAssign

denotes guided problem

denotes “paired problems” that develop reasoning with symbols and numerical values

The problems found in this chapter may be assigned online in Enhanced WebAssign1. denotes straightforward problem; 2. denotes intermediate problem; 3. denotes challenging problem 1. full solution available in the Student Solutions Manual/Study Guide

1. denotes problems most often assigned in Enhanced WebAssign; these provide students with targeted feedback and either a Master It tutorial or a Watch It solution video.

shaded

u

d

R

Figure P26.10

| Problems 765

21. Two capacitors give an equivalent capacitance of 9.00 pF when connected in parallel and an equivalent capacitance of 2.00 pF when connected in series. What is the capaci-tance of each capacitor?

22. Two capacitors give an equivalent capacitance of Cp when connected in parallel and an equivalent capacitance of Cs when connected in series. What is the capacitance of each capacitor?

23. Four capacitors are con-nected as shown in Figure P26.23. (a) Find the equivalent capacitance between points a and b. (b) Calculate the charge on each capacitor, tak-ing DVab 5 15.0 V.

24. Consider the circuit shown in Figure P26.24, where C1 5 6.00 mF, C2 5 3.00 mF, and DV 5 20.0 V. Capacitor C1 is first charged by closing switch S1. Switch S1 is then opened, and the charged capacitor is connected to the uncharged capacitor by closing S2. Calculate (a) the initial charge acquired by C1 and (b) the final charge on each capacitor.

25. Consider three capacitors C1, C2, and C3 and a battery. If only C1 is connected to the battery, the charge on C1 is 30.8 mC. Now C1 is disconnected, discharged, and con-nected in series with C2. When the series combination of C2 and C1 is connected across the battery, the charge on C1 is 23.1 mC. The circuit is disconnected, and both capaci-tors are discharged. Next, C3, C1, and the battery are con-nected in series, resulting in a charge on C1 of 25.2 mC. If, after being disconnected and discharged, C 1, C2, and C3 are connected in series with one another and with the bat-tery, what is the charge on C1?

26. Why is the following situation impossible? A technician is test-ing a circuit that contains a capacitance C. He realizes that a better design for the circuit would include a capacitance 73C rather than C. He has three additional capacitors, each with capacitance C. By combining these additional capaci-tors in a certain combination that is then placed in par-allel with the original capacitor, he achieves the desired capacitance.

27. A group of identical capacitors is connected first in series and then in parallel. The combined capacitance in parallel is 100 times larger than for the series connection. How many capaci-tors are in the group?

28. (a) Find the equivalent capacitance between points a and b for the group of capacitors connected as shown in Figure P26.28. Take C1 5 5.00 mF, C2 5 10.0 mF, and C3 5 2.00 mF. (b) What charge is stored on C3 if the potential difference between points a and b is 60.0 V?

combination, (b) the potential difference across each capacitor, and (c) the charge on each capacitor.

15. Find the equivalent capacitance of a 4.20-mF capacitor and an 8.50-mF capacitor when they are connected (a) in series and (b) in parallel.

16. Given a 2.50-mF capacitor, a 6.25-mF capacitor, and a 6.00-V battery, find the charge on each capacitor if you connect them (a) in series across the battery and (b) in parallel across the battery.

17. According to its design specification, the timer circuit delaying the closing of an elevator door is to have a capaci-tance of 32.0 mF between two points A and B. When one circuit is being constructed, the inexpensive but durable capacitor installed between these two points is found to have capacitance 34.8 mF. To meet the specification, one additional capacitor can be placed between the two points. (a) Should it be in series or in parallel with the 34.8-mF capacitor? (b) What should be its capacitance? (c) What If? The next circuit comes down the assembly line with capaci-tance 29.8 mF between A and B. To meet the specification, what additional capacitor should be installed in series or in parallel in that circuit?

18. Find (a) the equivalent capacitance of the capacitors in Fig-ure P26.18, (b) the charge on each capacitor, and (c) the potential difference across each capacitor.

9.00 V

8.00 Fµ8.00 Fµ 2.00 Fµ

6.00 Fµ

� �

Figure P26.18

19. For the system of four capacitors shown in Figure P26.19, find (a) the equivalent capacitance of the system, (b) the charge on each capacitor, and (c) the potential difference across each capacitor.

20. Three capacitors are connected to a battery as shown in Figure P26.20. Their capaci-tances are C1 5 3C, C2 5 C, and C3 5 5C. (a) What is the equiva-lent capacitance of this set of capacitors? (b) State the rank-ing of the capacitors according to the charge they store from largest to smallest. (c) Rank the capacitors according to the potential differences across them from largest to smallest. (d) What If? Assume C3 is increased. Explain what hap-pens to the charge stored by each capacitor.

90.0 V

4.00 Fµ

6.00 Fµ

2.00 Fµ

3.00 Fµ

� �

Figure P26.19 Problems 19 and 54.

C2 C3

C1

�

�

Figure P26.20

6.00 mF

20.0 mF

3.00 mF15.0

a

mF

b

Figure P26.23

C1 C2

S2S1

V��

�

Figure P26.24

C2 C2

C1 C1

C2 C2

C3

b

a

Figure P26.28


36. A parallel-plate capacitor has a charge Q and plates of area A. What force acts on one plate to attract it toward the other plate? Because the electric field between the plates is E 5 Q /AP0, you might think the force is F 5 QE 5 Q 2/AP0. This conclusion is wrong because the field E includes con-tributions from both plates, and the field created by the positive plate cannot exert any force on the positive plate. Show that the force exerted on each plate is actually F 5 Q 2/2AP0. Suggestion: Let C 5 P0A/x for an arbitrary plate separation x and note that the work done in separating the two charged plates is W 5 e F dx.

37. Two identical parallel-plate capacitors, each with capacitance 10.0 mF, are charged to potential difference 50.0 V and then disconnected from the battery. They are then connected to each other in parallel with plates of like sign connected. Finally, the plate separation in one of the capacitors is doubled. (a) Find the total energy of the sys-tem of two capacitors before the plate separation is doubled. (b) Find the potential difference across each capacitor after the plate separation is doubled. (c) Find the total energy of the system after the plate separation is doubled. (d) Recon-cile the difference in the answers to parts (a) and (c) with the law of conservation of energy.

38. Two identical parallel-plate capacitors, each with capacitance C, are charged to potential difference DV and then disconnected from the battery. They are then con-nected to each other in parallel with plates of like sign con-nected. Finally, the plate separation in one of the capaci-tors is doubled. (a) Find the total energy of the system of two capacitors before the plate separation is doubled. (b) Find the potential difference across each capacitor after the plate separation is doubled. (c) Find the total energy of the system after the plate separation is doubled. (d) Recon-cile the difference in the answers to parts (a) and (c) with the law of conservation of energy.

39. Review. The circuit in Figure P26.39 consists of two identical, parallel metal plates connected to identical metal springs, a switch, and a 100-V battery. With the switch open, the plates are uncharged, are separated by a distance d 5 8.00 mm, and have a capacitance C 5 2.00 mF. When the switch is closed, the distance between the plates decreases by a factor of 0.500. (a) How much charge collects on each plate? (b) What is the spring constant for each spring?

40. Consider two conducting spheres with radii R1 and R2 separated by a distance much greater than either radius. A total charge Q is shared between the spheres. We wish to show that when the electric potential energy of the system has a minimum value, the potential difference between the spheres is zero. The total charge Q is equal to q1 1 q2, where q1 represents the charge on the first sphere and q2 the charge on the second. Because the spheres are very far apart, you can assume the charge of each is uniformly distributed over its surface. (a) Show that the energy associated with a single conducting sphere of radius R and charge q surrounded by a vacuum is U 5 keq 2/2R.

29. Find the equivalent capaci-tance between points a and b in the combination of capacitors shown in Figure P26.29.

Section 26.4 Energy Stored in a Charged Capacitor

30. (a) A 3.00-mF capacitor is con-nected to a 12.0-V battery. How much energy is stored in the capacitor? (b) Had the capacitor been connected to a 6.00-V battery, how much energy would have been stored?

31. A 12.0-V battery is connected to a capacitor, resulting in 54.0 mC of charge stored on the capacitor. How much energy is stored in the capacitor?

32. The immediate cause of many deaths is ventricular fibrilla-tion, which is an uncoordinated quivering of the heart. An electric shock to the chest can cause momentary paralysis of the heart muscle, after which the heart sometimes resumes its proper beating. One type of defibrillator (chapter- opening photo, page 740) applies a strong electric shock to the chest over a time interval of a few milliseconds. This device contains a capacitor of several microfarads, charged to several thousand volts. Electrodes called paddles are held against the chest on both sides of the heart, and the capacitor is discharged through the patient’s chest. Assume an energy of 300 J is to be delivered from a 30.0-mF capaci-tor. To what potential difference must it be charged?

33. As a person moves about in a dry environment, electric charge accumulates on the person’s body. Once it is at high voltage, either positive or negative, the body can discharge via sparks and shocks. Consider a human body isolated from ground, with the typical capacitance 150 pF. (a) What charge on the body will produce a potential of 10.0 kV? (b) Sensitive electronic devices can be destroyed by electro-static discharge from a person. A particular device can be destroyed by a discharge releasing an energy of 250 mJ. To what voltage on the body does this situation correspond?

34. Two capacitors, C1 5 18.0 mF and C2 5 36.0 mF, are connected in series, and a 12.0-V battery is connected across the two capacitors. Find (a) the equivalent capaci-tance and (b) the energy stored in this equivalent capac-itance. (c) Find the energy stored in each individual capacitor. (d) Show that the sum of these two energies is the same as the energy found in part (b). (e) Will this equality always be true, or does it depend on the number of capacitors and their capacitances? (f) If the same capaci-tors were connected in parallel, what potential difference would be required across them so that the combination stores the same energy as in part (a)? (g) Which capacitor stores more energy in this situation, C1 or C2?

35. Two capacitors, C1 5 25.0 mF and C2 5 5.00 mF, are con-nected in parallel and charged with a 100-V power sup-ply. (a) Draw a circuit diagram and (b) calculate the total energy stored in the two capacitors. (c) What If? What potential difference would be required across the same two capacitors connected in series for the combination to store the same amount of energy as in part (b)? (d) Draw a circuit diagram of the circuit described in part (c).

ba

6.0 mF

5.0 mF7.0 mF

4.0 mF

Figure P26.29

kkd

V

S

��

Figure P26.39

| Problems 767

coated paper. Each strip of foil and paper is 7.00 cm wide. The foil is 0.004 00 mm thick, and the paper is 0.025 0 mm thick and has a dielectric constant of 3.70. What length should the strips have if a capacitance of 9.50 3 1028 F is desired before the capacitor is rolled up? (Adding a second strip of paper and rolling the capacitor would effectively double its capacitance by allowing charge storage on both sides of each strip of foil.)

(b) Find the total energy of the system of two spheres in terms of q1, the total charge Q, and the radii R1 and R2. (c) To minimize the energy, differentiate the result to part (b) with respect to q1 and set the derivative equal to zero. Solve for q1 in terms of Q and the radii. (d) From the result to part (c), find the charge q2. (e) Find the potential of each sphere. (f) What is the potential difference between the spheres?

41. Review. A storm cloud and the ground represent the plates of a capacitor. During a storm, the capacitor has a potential difference of 1.00 3 108 V between its plates and a charge of 50.0 C. A lightning strike delivers 1.00% of the energy of the capacitor to a tree on the ground. How much sap in the tree can be boiled away? Model the sap as water initially at 30.08C. Water has a specific heat of 4 186 J/kg ? 8C, a boiling point of 1008C, and a latent heat of vaporization of 2.26 3 106 J/kg.

Section 26.5 Capacitors with Dielectrics

42. The voltage across an air-filled parallel-plate capaci-tor is measured to be 85.0 V. When a dielectric is inserted and completely fills the space between the plates as in Figure P26.42, the voltage drops to 25.0 V. (a) What is the dielectric constant of the inserted material? (b) Can you identify the dielectric? If so, what is it? (c) If the dielectric does not completely fill the space between the plates, what could you conclude about the voltage across the plates?

C0 C

Dielectric

VV0� �

a b

Figure P26.42

43. (a) How much charge can be placed on a capacitor with air between the plates before it breaks down if the area of each plate is 5.00 cm2? (b) What If? Find the maximum charge if polystyrene is used between the plates instead of air.

44. A supermarket sells rolls of aluminum foil, plastic wrap, and waxed paper. (a) Describe a capacitor made from such materials. Compute order-of-magnitude estimates for (b) its capacitance and (c) its breakdown voltage.

45. Determine (a) the capacitance and (b) the maximum potential difference that can be applied to a Teflon-filled parallel-plate capacitor having a plate area of 1.75 cm2 and a plate separation of 0.040 0 mm.

46. A commercial capacitor is to be constructed as shown in Figure P26.46. This particular capacitor is made from two strips of aluminum foil separated by a strip of paraffin-

Aluminum

Paper7.00 cm

Figure P26.46

47. A parallel-plate capacitor in air has a plate separation of 1.50 cm and a plate area of 25.0 cm2. The plates are charged to a potential difference of 250 V and disconnected from the source. The capacitor is then immersed in distilled water. Assume the liquid is an insulator. Determine (a) the charge on the plates before and after immersion, (b) the capacitance and potential difference after immersion, and (c) the change in energy of the capacitor.

48. Each capacitor in the combination shown in Figure P26.48 has a breakdown voltage of 15.0 V. What is the breakdown voltage of the combination?

20.0 mF

10.0 mF

20.0 mF

20.0 mF

20.0 mF

Figure P26.48

Section 26.6 Electric Dipole in an Electric Field

49. An infinite line of positive charge lies along the y axis, with charge density l 5 2.00 mC/m. A dipole is placed with its center along the x axis at x 5 25.0 cm. The dipole consists of two charges 610.0 mC separated by 2.00 cm. The axis of the dipole makes an angle of 35.08 with the x axis, and the positive charge is farther from the line of charge than the negative charge. Find the net force exerted on the dipole.

50. A small object with electric dipole moment pS is placed in a nonuniform electric field E

S5 E 1x 2 i^. That is, the field

is in the x direction, and its magnitude depends only on the coordinate x. Let u represent the angle between the dipole moment and the x direction. Prove that the net force on the dipole is

F 5 padEdxb cos u

acting in the direction of increasing field.

51. A small, rigid object carries positive and negative 3.50-nC charges. It is oriented so that the positive charge


of the four-plate system P1P2P3P4? (d) What is the charge on P4?

has coordinates (21.20 mm, 1.10 mm) and the negative charge is at the point (1.40 mm, 21.30 mm). (a) Find the electric dipole moment of the object. The object is placed in an electric field E

S5 17.80 3 103 i^ 2 4.90 3 103 j^ 2 N/C.

(b) Find the torque acting on the object. (c) Find the potential energy of the object–field system when the object is in this orientation. (d) Assuming the orientation of the object can change, find the difference between the maxi-mum and minimum potential energies of the system.

Section 26.7 An Atomic Description of Dielectrics

52. The general form of Gauss’s law describes how a charge creates an electric field in a material, as well as in vacuum:

3ES? dA

S5

q in

P

where P 5 kP0 is the permittivity of the material. (a) A sheet with charge Q uniformly distributed over its area A is sur-rounded by a dielectric. Show that the sheet creates a uni-form electric field at nearby points with magnitude E 5 Q /2AP. (b) Two large sheets of area A, carrying opposite charges of equal magnitude Q , are a small distance d apart. Show that they create uniform electric field in the space between them with magnitude E 5 Q /2AP. (c) Assume the negative plate is at zero potential. Show that the positive plate is at potential Qd/AP. (d) Show that the capacitance of the pair of plates is given by C 5 AP/d 5 kAP0/d.

Additional Problems

53. Find the equivalent capacitance of the group of capacitors shown in Figure P26.53.

5.00 Fµ

4.00 Fµ

6.00 Fµ

3.00 Fµ

3.00 Fµ

7.00 Fµ

2.00 Fµ

� �

Figure P26.53

54. For the system of four capacitors shown in Figure P26.19, find (a) the total energy stored in the system and (b) the energy stored by each capacitor. (c) Compare the sum of the answers in part (b) with your result to part (a) and explain your observation.

55. Four parallel metal plates P1, P2, P3, and P4, each of area 7.50 cm2, are separated successively by a distance d 5 1.19 mm as shown in Figure P26.55. Plate P1 is connected to the negative terminal of a battery, and P2 is connected to the positive terminal. The battery maintains a potential difference of 12.0 V. (a) If P3 is connected to the negative terminal, what is the capacitance of the three-plate system P1P2P3? (b) What is the charge on P2? (c) If P4 is now con-nected to the positive terminal, what is the capacitance

12.0 V

P2 P3 P4P1

d d d

�

�

Figure P26.55

56. Two large, parallel metal plates, each of area A, are oriented horizontally and separated by a distance 3d. A grounded conducting wire joins them, and initially each plate carries no charge. Now a third identical plate carry-ing charge Q is inserted between the two plates, parallel to them and located a distance d from the upper plate as shown in Figure P26.56. (a) What induced charge appears on each of the two original plates? (b) What potential dif-ference appears between the middle plate and each of the other plates?

2d

d

Figure P26.56

57. A 2.00-nF parallel-plate capacitor is charged to an ini-tial potential difference DVi 5 100 V and is then isolated. The dielectric material between the plates is mica, with a dielectric constant of 5.00. (a) How much work is required to withdraw the mica sheet? (b) What is the potential dif-ference across the capacitor after the mica is withdrawn?

58. Why is the following situation impossible? A 10.0-mF capacitor has plates with vacuum between them. The capacitor is charged so that it stores 0.050 0 J of energy. A particle with charge 23.00 mC is fired from the positive plate toward the negative plate with an initial kinetic energy equal to 1.00 3 1024 J. The particle arrives at the negative plate with a reduced kinetic energy.

59. A parallel-plate capacitor is constructed using a dielec-tric material whose dielectric constant is 3.00 and whose dielectric strength is 2.00 3 108 V/m. The desired capaci-tance is 0.250 mF, and the capacitor must withstand a maxi-mum potential difference of 4.00 kV. Find the minimum area of the capacitor plates.

60. A parallel-plate capacitor with vacuum between its horizontal plates has a capacitance of 25.0 mF. A noncon-ducting liquid with dielectric constant 6.50 is poured into the space between the plates, filling up a fraction f of its volume. (a) Find the new capacitance as a function of f. (b) What should you expect the capacitance to be when f 5 0? Does your expression from part (a) agree with your

| Problems 769

(a), what will be the maximum voltage across each of the capacitors used?

65. A capacitor of unknown capacitance has been charged to a potential difference of 100 V and then disconnected from the battery. When the charged capacitor is then connected in parallel to an uncharged 10.0-mF capacitor, the poten-tial difference across the combination is 30.0 V. Calculate the unknown capacitance.

66. Example 26.1 explored a cylindrical capacitor of length , with radii a and b for the two conductors. In the What If? section of that example, it was claimed that increasing , by 10% is more effective in terms of increasing the capac-itance than increasing a by 10% if b . 2.85a. Verify this claim mathematically.

67. Capacitors C1 5 6.00 mF and C2 5 2.00 mF are charged as a parallel combination across a 250-V battery. The capacitors are disconnected from the battery and from each other. They are then connected positive plate to negative plate and negative plate to positive plate. Calculate the resulting charge on each capacitor.

68. A parallel-plate capacitor of plate separation d is charged to a potential difference DV0. A dielectric slab of thickness d and dielectric constant k is introduced between the plates while the battery remains connected to the plates. (a) Show that the ratio of energy stored after the dielectric is introduced to the energy stored in the empty capacitor is U/U0 5 k. (b) Give a physical explanation for this increase in stored energy. (c) What happens to the charge on the capacitor? Note: This situation is not the same as in Example 26.5, in which the battery was removed from the circuit before the dielectric was introduced.

Challenge Problems

69. Some physical systems possessing capacitance continu-ously distributed over space can be modeled as an infinite array of discrete circuit elements. Examples are a micro-wave waveguide and the axon of a nerve cell. To practice analysis of an infinite array, determine the equivalent capacitance C between terminals X and Y of the infinite set of capacitors represented in Figure P26.69. Each capacitor has capacitance C0. Suggestions: Imagine that the ladder is cut at the line AB and note that the equivalent capacitance of the infinite section to the right of AB is also C.

answer? (c) What capacitance should you expect when f 5 1? Does the expression from part (a) agree with your answer?

61. A 10.0-mF capacitor is charged to 15.0 V. It is next connected in series with an uncharged 5.00-mF capaci-tor. The series combination is finally connected across a 50.0-V battery as diagrammed in Figure P26.61. Find the new potential differences across the 5.00-mF and 10.0-mF capacitors after the switch is thrown closed.

62. (a) Two spheres have radii a and b, and their centers are a distance d apart. Show that the capacitance of this system is

C 54pP0

1a

11b

22d

provided d is large compared with a and b. Suggestion: Because the spheres are far apart, assume the potential of each equals the sum of the potentials due to each sphere. (b) Show that as d approaches infinity, the above result reduces to that of two spherical capacitors in series.

63. Two square plates of sides , are placed parallel to each other with separation d as suggested in Figure P26.63. You may assume d is much less than ,. The plates carry uni-formly distributed static charges 1Q 0 and 2Q 0. A block of metal has width ,, length ,, and thickness slightly less than d. It is inserted a distance x into the space between the plates. The charges on the plates remain uniformly dis-tributed as the block slides in. In a static situation, a metal prevents an electric field from penetrating inside it. The metal can be thought of as a perfect dielectric, with k S .̀ (a) Calculate the stored energy in the system as a function of x. (b) Find the direction and magnitude of the force that acts on the metallic block. (c) The area of the advanc-ing front face of the block is essentially equal to ,d. Consid-ering the force on the block as acting on this face, find the stress (force per area) on it. (d) Express the energy density in the electric field between the charged plates in terms of Q 0, ,, d, and P0. (e) Explain how the answers to parts (c) and (d) compare with each other.

5.00 mF

50.0 V

Vi � 15.0 V

��10.0 mF

�

� �

Figure P26.61

xd

�

� � � � �

� � � � �

�Q 0

�Q 0

Figure P26.63

64. To repair a power supply for a stereo amplifier, an electronics technician needs a 100-mF capacitor capable of withstanding a potential difference of 90 V between the plates. The immediately available supply is a box of five 100-mF capacitors, each having a maximum voltage capa-bility of 50 V. (a) What combination of these capacitors has the proper electrical characteristics? Will the technician use all the capacitors in the box? Explain your answers. (b) In the combination of capacitors obtained in part

C0

C0

C0

X

Y

A

B

Figure P26.69

70. The inner conductor of a coaxial cable has a radius of 0.800 mm, and the outer conductor’s inside radius is 3.00 mm. The space between the conductors is filled with polyethylene, which has a dielectric constant of 2.30 and


a common value k, your result becomes the same as the capacitance of a capacitor containing a single dielectric: C 5 kP0LW/t.

73. Calculate the equivalent capacitance between points a and b in Figure P26.73. Notice that this system is not a simple series or parallel combination. Suggestion: Assume a poten-tial difference DV between points a and b. Write expres-sions for DVab in terms of the charges and capacitances for the various possible pathways from a to b and require con-servation of charge for those capacitor plates that are con-nected to each other.

a dielectric strength of 18.0 3 106 V/m. What is the maxi-mum potential difference this cable can withstand?

71. A capacitor is constructed from two square, metallic plates of sides , and separation d. Charges 1Q and 2Q are placed on the plates, and the power supply is then removed. A material of dielectric constant k is inserted a distance x into the capacitor as shown in Figure P26.71. Assume d is much smaller than x. (a) Find the equivalent capacitance of the device. (b) Calculate the energy stored in the capacitor. (c) Find the direction and magnitude of the force exerted by the plates on the dielectric. (d) Obtain a numerical value for the force when x 5 ,/2, assuming , 5 5.00 cm, d 5 2.00 mm, the dielectric is glass (k 5 4.50), and the capacitor was charged to 2.00 3 103 V before the dielectric was inserted. Suggestion: The system can be considered as two capacitors connected in parallel.

xd

�

� � � � �

� � � � �

�Q

�Q

k

Figure P26.71

72. A parallel-plate capacitor with plates of area LW and plate separation t has the region between its plates filled with wedges of two dielectric materials as shown in Figure P26.72. Assume t is much less than both L and W. (a) Determine its capacitance. (b) Should the capacitance be the same if the labels k1 and k2 are interchanged? Dem-onstrate that your expression does or does not have this property. (c) Show that if k1 and k2 approach equality to

k2k1t

LW

Figure P26.72

C

C

3C

2C

2C

Figure P26.75

a

b2.00 mF

4.00 mF

2.00 mF 4.00 mF8.00 mF

Figure P26.73

74. Consider two long, parallel, and oppositely charged wires of radius r with their centers separated by a distance D that is much larger than r. Assuming the charge is dis-tributed uniformly on the surface of each wire, show that the capacitance per unit length of this pair of wires is

C,

5pP0

ln 1D/r 2 75. Determine the equivalent capacitance of the combina-

tion shown in Figure P26.75. Suggestion: Consider the sym-metry involved.

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